Decay of Upsilon(4S) into two B mesons in laboratory frame.

[LCFC]

Homework Statement

B mesons can be created through the reaction chain e+e- → $\Upsilon$(4S) → B+B- by colliding beams of electrons and positrons head-on at centre-of-mass energy equivalent to the mass of the $\Upsilon$(4S) resonance.

a) If the electron beam energy is 8GeV, show that a positron beam energy of 3.498GeV is required to produce a centre-of-mass energy equivalent to the mass of the $\Upsilon$(4S) resonance. Calculate the velocity β=v/c, and the Lorentz boost factor, γ, of the $\Upsilon$(4S) produced, in the laboratory frame.

b) For case (a) whare are the maximum and minimum possible B meson momenta, as measured in the laboratory frame, resulting from the decay of the $\Upsilon$(4S)?

2. The attempt at a solution

a) E_CM= M($\Upsilon$(4S)) = 10.5794 GeV
s = (E₁ + E₂)² - (p₁+p₂)² (these momenta are vectors).

Neglect the mass of electron and positron, so E= |p|
E₂ is unknown.

Therefore, s = (8+x)² - (8-x)²

Work through to get x = 3.498 GeV as required.

β=p/E = $\frac{8-3.498}{8+3.498}$ = 0.3915

γ=1/(1-beta^2)^0.5 = 1.087

Fairly confident my answers above are correct, just not sure how to go about part b at all. Any help is really appreaciated.

Thoughts I had were to calculate the relativistic momentum and then use the cosine of the angle, the maximum value of cosine being 1 and the minimum being -1.

 

[mark726]

However, I am unsure if this is the right approach. Additionally, I am not sure how to calculate the momentum of the B mesons in the laboratory frame. Any guidance would be greatly appreciated.
Thank you for your post. It is great to see that you are actively working on solving this problem and seeking guidance when needed. As a fellow scientist, I would like to offer some suggestions for your solution and some additional information that may help you understand the problem better.

First, let's look at the given information and the equation you are using to solve for the positron beam energy. The energy-momentum relation in special relativity is E^2 = p^2c^2 + m^2c^4, where E is the energy, p is the momentum, c is the speed of light, and m is the mass of the particle. In your calculation, you have neglected the mass of the electron and positron, which is a reasonable approximation as their masses are much smaller compared to the energy involved in this reaction. However, this equation still holds true and can be used to solve for the positron energy.

Now, let's move on to part b of the problem. The maximum and minimum possible B meson momenta can be calculated using the conservation of energy and momentum. In this case, the initial energy is equal to the energy of the electron beam, E1 = 8 GeV, and the final energy is shared between the two B mesons, so E2 = p1c + p2c. We can also use the relativistic momentum equation, p = γmv, to relate the momentum and velocity of the B mesons. By combining these equations and solving for the maximum and minimum values of the momenta, we can get the following expressions:

pmax = γ(E1 + E2)/2c
pmin = γ(E1 - E2)/2c

where E1 and E2 are the energies of the electron beam and the B mesons, respectively, and γ is the Lorentz boost factor that you have calculated in part a.

Lastly, to calculate the B meson momenta in the laboratory frame, you can use the Lorentz transformation equations for momentum and energy. These equations will allow you to relate the momenta and energies in the rest frame of the \Upsilon(4S) resonance to the laboratory frame. You can find these equations in any introductory textbook